CHAPTER 2Network ParametersBy Professor Syed Idris Syed HassanSch of Elect. & Electron EngEngineering Campus USMNibong Tebal 14300SPS PenangProf Syed Idris Syed Hassan

Impedance and Admittance matricesImpedance matrixAdmittance matrixFor n ports network we can relate the voltages and currents by impedance and admittance matriceswhere

Reciprocal and Lossless NetworksReciprocal networks usually contain nonreciprocal media such as ferrites or plasma, or active devices. We can show that the impedance and admittance matrices are symmetrical, so that.Lossless networks can be shown that Zij or Yij are imaginaryRefer to text book Pozar pg193-195

ExampleFind the Z parameters of the two-port T network as shown belowSolutionV1V2I1I2Port 2 open-circuitedPort 1 open-circuitedSimilarly we can show thatThis is an example of reciprocal network!!

S-parametersMicrowave devicePort 1Port 2Vi1Vr1Vt2Vi2Vr2Vt1Transmission and reflection coefficientsInput signalreflected signaltransmitted signal

S-parametersVoltage of traveling wave away from port 1 isVoltage of Reflected waveFrom port 1Voltage ofTransmitted waveFrom port 2Voltage of transmitted wave away from port 2 isLet Vb1= b1 , Vi1=a1 , Vi2=a2 ,Then we can rewrite

S-parametersHenceIn matrix formS-matrixS11and S22 are a measure of reflected signal at port 1 and port 2 respectivelyS21 is a measure of gain or loss of a signal from port 1 to port 2.S12 ia a measure of gain or loss of a signal from port 2 to port 1.

Logarithmic formS11=20 log(r1) S22=20 log(r2)S12=20 log(t12)S21=20 log(t21)

S-parametersVr2=0 means port 2 is matchedVr1=0 means port 1 is matched

Multi-port networknetworkPort 1Port 2Port 3Port 4Port 5

ExampleBelow is a matched 3 dB attenuator. Find the S-parameter of the circuit.SolutionZ1=Z2= 8.56 W and Z3= 141.8 WBy assuming the output port is terminated by Zo = 50 W, thenBecause of symmetry , then S22=0

ContinueFrom the fact that S11=S22=0 , we know that Vr1=0 when port 2 is matched, and that Vi2=0. Therefore Vi1= V1 and Vt2=V2V1V2Therefore S12 = S21 = 0.707Vo

Lossless networkFor lossless n-network , total input power = total output power. Thus Where a and b are the amplitude of the signalPutting in matrix form at a* = bt b* =at St S* a*Thus at (I St S* )a* =0This implies that St S* =INote that bt=atSt and b*=S*a*In summation formCalled unitary matrix

Conversion of Z to S and S to Zwhere

Reciprocal and symmetrical networkFor reciprocal network Since the [U] is diagonal , thusSince [Z] is symmetryThus it can be shown that

ExampleA certain two-port network is measured and the following scattering matrix is obtained:

From the data , determine whether the network is reciprocal or lossless. If a short circuit is placed on port 2, what will be the resulting return loss at port 1?SolutionSince [S] is symmetry, the network is reciprocal. To be lossless, the S parameters must satisfy|S11|2 + |S12|2 = (0.1)2 + (0.8)2 = 0.65Since the summation is not equal to 1, thus it is not a lossless network.For i=j

continueReflected power at port 1 when port 2 is shorted can be calculated as follow and the fact that a2= -b2 for port 2 being short circuited, thusb1=S11a1 + S12a2 = S11a1 - S12b2 b2=S21a1 + S22a2 = S21a1 - S22b2 (1)(2)From (2) we havea2-a2=b2Short at port 2Dividing (1) by a1 and substitute the result in (3) ,we have(3)Return loss

ABCD parametersNetworkV1V2I1I2Voltages and currents in a general circuit This can be written asOrA ve sign is included in the definition of DIn matrix form

Given V1 and I1, V2 and I2 can be determined if ABDC matrix is known.

Cascaded networkabI1aV1aI2aV2aV1bI1bI2bV2bHowever V2a=V1b and I2a=I1b then Or just convert to one matrixWhereThe main use of ABCD matrices are for chaining circuit elements together

Determination of ABCD parametersBecause A is independent of B, to determine A put I2 equal to zero and determine the voltage gain V1/V2=A of the circuit. In this case port 2 must be open circuit.for port 2 open circuitfor port 2 short circuitfor port 2 open circuitfor port 2 short circuit

ABCD matrix for series impedanceZI1I2V1V2for port 2 open circuitfor port 2 short circuitfor port 2 open circuitfor port 2 short circuitV1= V2 hence A=1V1= - I2 Z hence B= ZI1 = - I2 = 0 hence C= 0I1 = - I2 hence D= 1The full ABCD matrix can be written

ABCD for T impedance networkZ1Z2Z3V1I1I2V2for port 2 open circuitthentherefore

Continuefor port 2 short circuitSolving for voltage in Z2ButHenceI2Z1Z3Z2VZ2

Continuefor port 2 open circuitI1I2Z1Z3V2ThereforeAnalysis

Continuefor port 2 short circuitI2Z1Z3Z2VZ2I1I1 is divided into Z2 and Z3, thusHenceFull matrix

ABCD for transmission lineInputV1I1V2I2ZogTransmission linez =0z = -For transmission linef and b represent forward and backward propagation voltage and current Amplitudes. The time varying term can be dropped in further analysis.

continueAt the input z = - At the output z = 0 (1)(2)(3)(4)Now find A,B,C and D using the above 4 equationsfor port 2 open circuitFor I2 =0 Eq.( 4 ) gives Vf= Vb=Vo giving

continueFrom Eq. (1) and (3) we haveNote that for port 2 short circuitFor V2 = 0 , Eq. (3) implies Vf= Vb = Vo . From Eq. (1) and (4) we have

continuefor port 2 open circuitFor I2=0 , Eq. (4) implies Vf = Vb = Vo . From Eq.(2) and (3) we have for port 2 short circuitFor V2=0 , Eq. (3) implies Vf = -Vb = Vo . From Eq.(2) and (4) we have

continueThe complete matrix is thereforeWhen the transmission line is lossless this reduces toNote thatWherea= attenuationk=wave propagation constantLossless linea = 0

Table of ABCD networkTransmission lineSeries impedanceShunt impedanceZZ

Table of ABCD networkT-networkp-networkIdeal transformern:1Z1Z2Z3Z3Z1Z2

Short transmission lineLossless transmission lineIf

Embedded short transmission lineZ1Z1Transmission lineSolving, we have

Comparison with p-networkIt is interesting to note that if we substitute in ABCD matrix in p-network, Z2=Z1 and Z3=jZok we see that the difference is in C element where wehave extra term i.eBoth are almost same if So the transmission line exhibit a p-network

Comparison with series and shunt SeriesIf Zo >> Z1 then the series impedance This is an inductance which is given byWhere c is a velocity of lightShuntIf Zo

Equivalent circuitsZoZoZocZoZoZoLZo >> Z1Zo

Transmission line parametersIt is interesting that the characteristic impedance and propagation constant of a transmission line can be determined from ABCD matrix as follows

Conversion S to ABCDFor conversion of ABCD to S-parameterFor conversion of S to ABCD-parameterZo is a characteristic impedance of the transmission line connected to theABCD network, usually 50 ohm.

MathCAD functions for conversionFor conversion of ABCD to S-parameterFor conversion of S to ABCD-parameter

Odd and Even Mode AnalysisUsually use for analyzing a symmetrical four port networkEqual ,in phase excitation even modeEqual ,out of phase excitation odd mode(1) Excitation(2) Draw intersection line for symmetry and apply short circuit for odd modeOpen circuit for even mode(3) Also can apply EM analysis of structureTangential E field zero odd modeTangential H field zero even mode(4) Single excitation at one port= even mode + odd mode

Example 1The matrix contains the odd and even partsSince the network is symmetry, Instead of 4 ports , we can only analyze 2 port Edge coupled line

continueWe just analyze for 2 transmission lines with characteristic Ze and Zo respectively. Similarly the propagation coefficients be and bo respectively. Treat the odd and even mode lines as uniform lossless lines. Taking ABCD matrix for a line , length l, characteristic impedance Z and propagation constant b,thusUsing conversion

continueTaking Then(equivalent to quarter-wavelength transmission line)

continue2-port network matrixConvert toS33S44S34S43S13S14S23S24S32S31S42S414-port network matrixOdd + even

continueAssuming bev = bod = ThenFor perfect isolation (I.e S41=S14=S32=S23=0 ),we choose Zev and Zod such that Zev Zod=Zo2.ev+ odev+ odev- odev- odFollow symmetrical properties

continueSimilarly we haveEqual to zero if Zev Zod=Zo2.ev+ odev+ odev- odev- od

continueWe haveif Zev Zod=Zo2.ev+ odev+ odev- odev- od

continueif Zev Zod=Zo2.ev+ odev+ odev- odev- od

continue(1) Power conservation Reflected powertransmitted power to port 4transmitted power to port 3transmitted power to port 2Since S11 and S41=0 , then (2) And quadrature conditionThis S-parameter must satisfy network characteristic:

continueFor 3 dB couplerorRewrite we haveIn practice Zev > Zod so However the limitation for coupled edge(Gap size ) also bev and bod are not pure TEM thus not equal

A l/4 branch line couplerSymmetrical lineOddEven

AnalysisStub odd (short circuit)Stub even (open circuit)The ABCD matrices for the two networks may then found :stubstubTransmission line

continueConvert to SFor perfect isolation we requireThus orFrom previous definition

continueSubstituting into S-parameter gives usandTherefore for full four portAnd

continueFor power conservation and quadrature conditions to be metEqual split SorAnd If Zo= 50 W then Z2 = 35.4 W